# Why does the annihilation operator acting on the ground state in Quantum Field Theory gives a zero?

One of the main motivations for Quantum Field Theory after Dirac Equation is that the Dirac equation predicts negative energy states which leads to the ground state being unbounded which ultimately leads to the sea of negative energy electrons. It is said that the field theoretic approach cures it of this problem.

But every QFT book I have seen often defines the ground state $|0\rangle$ as:

$$a|0\rangle=0$$

but I can't see the basis for this definition ?

Of course, once I define the ground state as this it will automatically never lead to negative energy states? Why isn't this arbitrary? Why could I not do the same with Dirac Equation and define its ground state as such and hence remove all such negative energy states?

• – mpv Jun 12 '17 at 9:21

## 1 Answer

For the Klein Gordon equation, when we interpret it as being the equation for a wavefunction of a particle, we have the negative energy states as a solution.

This problem vanishes when we move to field theory and think of KG eqn as the differential equation for a classical field. Write the Hamiltonian for it and it is positive definite. Write the Hamiltonian in terms of ladder operators and require the condition that norm of the states must be positive. You will see that the eignevalues of the number operator must be positive. Hence, the problem for -ve energy states is solved for KG by interpreting it as the equation for a field.

The Dirac equation, in a sense is already a quantum field and the negative energy states thus, cannot be eliminated. Well, actually you can.

By the way, Majorana modified Dirac's equation to include only positive energy states. There is an article by Frank Wilczek of MIT on this topic.